Find the minimum value of #log_2^2 2 + log_ 2^3 2^2 + log_ 2^4 2^3.............log_2^n 2^(n-1)#?

Question

Find the minimum value of

#log_(2^2) 2 + log_ (2^3) 2^2 + log_ (2^4) 2^3.............log_(2^n) 2^(n-1)#

Eva Adamson · Accepted Answer

The Reqd. Minimum Value =#(n-1)(1/n)^(1/(n-1)).#

We know the Change of Base Rule : #log_ca=log_ba/log_bc.#

Using #b!=1# as common base to all #log,# we have,

The Expression#=log_4 2+log_8 4+log_(16)8+...+log_(2^n) 2^(n-1),#

#=log_b2/log_b4+log_b4/log_b8+log_b8/log_b 16+...+log_b 2^(n-1)/log_b(2^n),#

#=log_b2/log_b2^2+log_b2^2/log_b2^3+log_b2^3/log_b 2^4+...+log_b 2^(n-1)/log_b(2^n),#

#=log_b2/(2log_b2)+(2log_b2)/(3log_b2)+(3log_b2)/(4log_b 2)+...+((n-1)log_b 2)/(nlog_b2),#

#=1/2+2/3+3/4+...+(n-1)/n....(ast)#

Now, we use the Arithmetic Mean-Geometric Mean Inequality, i.e.,

#AM ge GM#

Observe that the AM of #(n-1) +ve" Nos. of "(ast)# is,

#1/(n-1){1/2+2/3+3/4+...+(n-1)/n},# and, their GM is,

#{(1/2)(2/3)(3/4)...((n-1)/n)}^(1/(n-1))=(1/n)^(1/(n-1)).#

#:., 1/(n-1){1/2+2/3+3/4+...+(n-1)/n} ge (1/n)^(1/(n-1)), i.e.,#

# {1/2+2/3+3/4+...+(n-1)/n} ge (n-1)(1/n)^(1/(n-1)).#

Therefore, the Reqd. Minimum Value =#(n-1)(1/n)^(1/(n-1)).#

Enjoy Maths.!

Eva Abramson · Answer

#sum_(k=1)^n log_(2^k) 2^(k-1) = n+sum_(k=1)^n(-1)^k1/k((n),(k))#

#log_(a^b) a^c = log a^c/(log a^b)=c/b#

so

#sum_(k=1)^n log_(2^k) 2^(k-1)= sum_(k=1)^n (k-1)/k = n-sum_(k=1)^n 1/k#

we know that

#1+x+x^2+ cdots + x^(n-1) = (x^n-1)/(x-1)#

so

#int_0^1 (x^n-1)/(x-1) dx = -int_0^1 ((1-y)^n-1)/y dy = - sum_(k=1)^n(-1)^k1/k((n),(k))#

so finally

#sum_(k=1)^n log_(2^k) 2^(k-1) = n+sum_(k=1)^n(-1)^k1/k((n),(k))#

Hazel Anglin · Answer

undefined The minimum value of the expression log₂² 2 + log₂³ 2² + log₂⁴ 2³ + ... + log₂ⁿ 2^(n-1) occurs when each term is minimized. Since log₂(x) is a monotonically increasing function, the minimum value of each term occurs when the argument (2, 2², 2³, ..., 2^(n-1)) is minimized, which is 2. Therefore, the minimum value of the expression is obtained when each term equals log₂ 2, and there are n terms. Hence, the minimum value is n * log₂ 2, which simplifies to n.

HIX Tutor · Answer

undefined To find the minimum value of the expression, let's break it down step by step.

Step 1: Simplify each term.
- The term log_2^2 2 means the logarithm to the base 2 of 2 squared, which simplifies to 2.
- Similarly, log_2^3 2^2 simplifies to 3.
- Continuing this pattern, log_2^4 2^3 simplifies to 4, and so on, until log_2^n 2^(n-1) which simplifies to n.

Step 2: Add up all the simplified terms.
We now have 2 + 3 + 4 + ... + n.

Step 3: Find the sum of the sequence.
To find the sum of this sequence, we can use a formula.
The sum of a sequence of consecutive numbers from 1 to n can be found using the formula: sum = (n/2) * (first term + last term).

In this case, the first term is 2 and the last term is n, so the sum of the sequence is:
sum = (n/2) * (2 + n).

Step 4: Find the minimum value.
To find the minimum value of this expression, we need to minimize the sum. We can do this by finding the value of n that makes the sum as small as possible.

Since we want to minimize the sum, we set the derivative of the sum with respect to n equal to 0:
d(sum)/dn = 0.

Simplifying and solving for n gives us:
n/2 + 1/2 = 0,
n/2 = -1/2,
n = -1.

However, since n represents the number of terms in the sequence, it cannot be negative.
Thus, we conclude that the minimum value of the expression is when n = 1, which leads to the sum being 3.

Therefore, the minimum value of the given expression is 3.

Find the minimum value of #log_2^2 2 + log_ 2^3 2^2 + log_ 2^4 2^3.............log_2^n 2^(n-1)#?

Find the minimum value of #log_(2^2) 2 + log_ (2^3) 2^2 + log_ (2^4) 2^3.............log_(2^n) 2^(n-1)#

Find the minimum value of

#log_(2^2) 2 + log_ (2^3) 2^2 + log_ (2^4) 2^3.............log_(2^n) 2^(n-1)#